If $\vert u \vert <1$ then the series converges the comparison test (the coefficients $a_k$ are bounded). On the other hand, if $\vert u \vert>1$ the series diverges by the divergence test (the coefficients decay slower than any exponential as we show below). Hence, we are left to consider what happens for $\vert u \vert=1$.
It seems to be false that for $\vert u \vert =1$ the series diverges (for $u\neq \pm i)$. It is unclear to me whether the series converges if $u=\pm i$.
Warm-up: u=1
Pick for simplicity $u=1$, then we are interested in the convergence of
$$ \sum_{k=1}^\infty a_k. $$
To show that $\sum_{k=1}^\infty a_k$ converges, it is enough to show that both
$$ \sum_{k=1}^\infty a_{2k}, \quad \sum_{k=1}^\infty a_{2k+1} $$
converge. I will focus on the first term, the second can be dealt with in a similar fashion.
We will show that $\sum_{k=1}^\infty a_{2k}$ satisfies the alternating series test and hence converges.
For this we note that
\begin{align*}
\frac{-k(k+1)+\lambda}{(k+2)(k+1)} = \frac{-1-\frac{1}{k}+ \frac{\lambda}{k^2}}{1+\frac{3}{k}+\frac{2}{k^2}}.
\end{align*}
Now we expand the denominator and get
\begin{align*}
\frac{-1-\frac{1}{k}+ \frac{\lambda}{k^2}}{1+\frac{3}{k}+\frac{2}{k^2}}
&= \left( -1-\frac{1}{k}+ \frac{\lambda}{k^2}\right) \left( 1-\frac{3}{k}+O(1/k^2)\right) \\
&= -1 + \frac{2}{k} + O(1/k^2).
\end{align*}
Hence, we get (for $k$ large enough)
$$ \vert a_{k+2} \vert < \vert a_k \vert. $$
Furthermore, we can write for some $K$ large enough
$$ a_{2n+K} = a_K(-1)^n \underbrace{\prod_{j=0}^{n-1} \frac{(-\lambda+(K+2j)(K+2j+1)}{(K+2j+2)(K+2j+1)}}_{=: b_{2n+K}}.$$
We have $b_{2n+K}>0$. So the only thing that is left to check is that $\lim_{n\rightarrow \infty} b_{2n+K}=0$. It is a general fact that an infinite product of terms in $(0,1)$ converges to zero iff the series over the logarithms of the factors diverges (see How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$? and the links therein). In our case we have to deal with
$$ \prod_{j=0}^\infty \left( 1- \frac{2}{j} + O(1/j^2) \right).$$
We check that
$$ \sum_{j=0}^\infty \left(\frac{2}{j} + O(1/j^2) \right) = \infty $$
as $ \sum_{j=0}^\infty O(1/j^2) $
converges (by the comparison test with $C/j^2$) and the fact that the harmonic series diverges to $+\infty$. Thus,
$$ \lim_{n\rightarrow \infty} b_{2n+K} = \prod_{j=0}^\infty \frac{(-\lambda+(K+2j)(K+2j+1)}{(K+2j+2)(K+2j+1)} = \prod_{j=0}^\infty \left(1 - \frac{2}{K+2j} + O(1/j^2)\right)= 0. $$
General $\vert u \vert =1$ where $u\neq \pm i$:
For a more general $u\in \mathbb{C}\setminus\{\pm i\}$ with $\vert u \vert=1$ we can resort to the Dirichlet test (see Explanation for the Proof of Dirichlet's test). In our case we have rewrite our series as
$$ \sum_{k=1}^\infty a_{2k+K} u^{2k+K} = \sum_{k=1} (-1)^k u^{2k+K} b_{2k+K}. $$
We have already shown that $(b_{2k+K})_{k\geq 1}$ is monotone decreasing and $\lim_{k\rightarrow \infty} b_{2k+K}=0$. Thus, we are left to check that there exists $M>0$ such that for all $N\in \mathbb{N}$ we have
$$ \left\vert \sum_{k=1}^N (-1)^k u^{2k+K} \right\vert \leq M $$
We note that that the series in question is a geometric series and we have (for $u^2\neq -1,$ i.e. $u\neq \pm i$)
$$ \sum_{k=1}^N (-1)^k u^{2k+K} = u^K \sum_{k=1}^N (-u^2)^k = -u^{K+2} \frac{1-(-u^2)^N}{1+u^2}. $$
Therefore, we have for $\vert u \vert =1$ and $u\neq \pm i$
$$ \left\vert \sum_{j=1}^N (-1)^k u^{2k+K} \right\vert \leq \frac{2}{\vert 1+u^2\vert}. $$
Hence,
$$ \sum_{k=1}^\infty a_{2k} u^{2k} $$
converges by Dirichlet's criterion.
Special case $u=\pm i$:
I do not see on the spot whether the series converges in this case. I would assume here it starts to depend on the relation between $a_0$ and $a_1$. Maybe if we choose $a_0$ and $a_1$ in a suitable way we can use again Dirichlet's test with $\vert\sum_{j=1}^N (\pm i)^j \vert \leq \sqrt{2}.$
@Severin: I think that it would make more sense to |u|=1, but in the problem it is written |u| --> 1 (the professor wrote this problem by hand and is a Physics professor (not Math))
– logpiy Apr 18 '23 at 18:39